Unit - 1
Properties of Fluid
Q1) What is Fluid? Explain in detail the definition of Fluid Machine?
A1) Fluid:
A fluid may be defined as follows:
“A fluid is a substance which is capable of flowing.”
Fluid Mechanics:
Fluid Mechanics may be defined as that branch of engineering Science which deals with the behavior of fluid under the conditions of rest and motion. The fluid mechanics may be divided into three parts: Statics, Kinematics and dynamics.
Statics: The study of incompressible fluids under static conditions is called hydrostatics and that dealing with the compressible static gases is termed as aerostatics.
Kinematics: It deals with the velocities, accelerations and the patterns of flow only. Forcesor energy causing velocity and acceleration are not dealt under this heading.
Dynamics: It deals with the relations between velocities accelerations of fluids with the forces or energy causing them.
Q2) Explain in detail Properties of fluid?
A2) Density or Mass Density:
Density or mass density of a fluid is defined as the ratio of the mass of a fluid to its volume.
It is denoted by the symbol (rho).
The unit of mass density in SI unit is kg/m3.
= Mass of fluid / Volume of fluid = m/v
The value of density of water is 1000 kg/m3
Specific weight or weight density
Specific weight or weight density of a fluid is the ratio between the weight of a fluid to its volume.
It is denoted by the symbol w.
w = Weight of fluid / Volume of fluid= W/v
= x g
The value of specific weight or weight density(w) for water is 9810N/m3.
Specific Volume:
Specific Volume of a fluid is defined as the volume of a fluid occupied by a unit mass or volume per unit mass of a fluid is called specific volume.
Mathematically, it is expressed as,
Specific Volume = Volume of Fluid / Mass of Fluid= v/m
= 1/
It is expressed as m3 /kg.
Specific Gravity:
Specific Gravity is defined as the ratio of the weight density of a fluid to the weight density of standard fluid.
Specific gravity is also called relative density. It is dimensionless gravity and is denoted by the symbol S.
S (For Liquids) = Weight density(density) of liquid / Weight density(density) of
Water.
The specific gravity of mercury is 13.6 and specific gravity of water is 1.
Q3) What is Viscosity? Explain Newton’s law of viscosity?
A3) Viscosity may be defined as the property of fluid which determines its resistance to shearing stresses. It is a measure of the internal fluid friction which causes resistance to flow.
Viscosity of fluids is due to cohesion and interaction between particles.
du/dy
= du/dy
= Constant of proportionality and is know as co-efficiency of dynamic viscosity.
du/dy = Rate of shear stress or rate of shear deformation or velocity gradient.
We have,
= / [du/dy]
Thus viscosity may also be defined as the shear stress required to produce unit rate of shear strain.
Unit Of Viscosity:
In S.I. units: N.s/m2
Kinematic Viscosity:
Kinematic Viscosity is defined as the ratio between the dynamic viscosity and density of fluid. It is denoted by v (called nu).
v = Viscosity / Density
= /
Units Of Kinematic Viscosity:
In SI units: m2/s
Newton’s Law of Viscosity
This law states that the shear stress () on a fluid element layer is directly proportional to the rate of shear strain. The constant of proportionality is called the co-efficient of viscosity.
= du / dy
The fluids which follow this law are known as Newtonian fluids.
Q4) Explain the Rheological Diagram in detail.
A4)
Types of Fluids
Ideal Fluid: A fluid which is incompressible and is having no viscosity, is known as an ideal fluid. Ideal Fluid is only an imaginary fluids as all the fluids. which exist have some viscosity.
Real Fluid: A fluid, which possesses viscosity, is known as real fluid. All the fluids in actual practice are real fluids.
Newtonian Fluid: A real fluid, in which the shears stress is directly proportional to the rate of shear strain is known as a Newtonian fluid.
Eg. Water, Kerosine.
Non - Newtonian Fluid: A real fluid, in which the shear stress is not proportional to the rate of shear stain known as a Non – Newtonian Fluid.
Example Blood, mud.
Ideal Plastic Fluid: A fluid, in which shares stress is more than the yield value and shear stress is proportional to the rate of shear strain is known as ideal plastic fluid.
Q5) Differentiate the Dynamic and Kinematic viscosity?
A5) Viscosity may be defined as the property of fluid which determines its resistance to shearing stresses. It is a measure of the internal fluid friction which causes resistance to flow.
Viscosity of fluids is due to cohesion and interaction between particles.
du/dy
= du/dy
= Constant of proportionality and is known as co-efficiency of dynamic viscosity.
du/dy = Rate of shear stress or rate of shear deformation or velocity gradient.
We have,
= / [du/dy]
Thus viscosity may also be defined as the shear stress required to produce unit rate of shear strain.
Units of Viscosity:
In S.I. units : N.s/m2
Kinematic Viscosity:
Kinematic Viscosity is defined as the ratio between the dynamic viscosity and density of fluid. It is denoted by v (called nu).
v = Viscosity / Density
= /
Units of Kinematic Viscosity :
In SI units: m2/s.
Q6) Explain surface tension with the help of diagram, also define Cohesion and adhesion.
A6) Surface tension
Surface tension is caused by the force of cohesion at the free surface.
A liquid molecules in the interior of the liquid mass is surrounded by the other molecules all around and is in equilibrium.
At the free surface of the liquid, there are no liquid molecules above the surface to balance the force of the molecules below it.
Consequently, as shown in fig there is net inward force on the molecules at the free surface.
The force is normal to the liquid surface.
At the free surface a thin layer of molecules is formed.
It is denoted by letter .
It’s SI unit is N/m.
Cohesion: Cohesion means intermolecular attraction between molecules of same liquid. It enables a liquid to resist small amount of tensile stresses. Cohesion is a tendency of the liquid to remain as one assemblage of particles.
Adhesion: Adhesion means attraction between the molecule of a liquid and the molecule of a solid boundary surface is contact with the liquid. These properties enable a liquid to stick to another body.
Q7) Explain the concept Capillarity, Compressibility, bulk modulus, vapour pressure.
A7) Capillarity:
Capillarity is defined as a phenomenon of rise or fall of a liquid surface in small tube relative to the adjacent general level of liquid when the tube is held vertically in the liquid.
The rise of liquid surface is known as capillary rise while the fall of the liquid surface is known as capillary depression.
It is expressed in terms of cm or mm of liquid.
Its value depends upon the specific weight of the liquid, diameter of the tube and surface tension of the liquid.
h =Height of the liquid in the tube.
d = Diameter of tube
= Surface tension of liquid.
= Angle of contact between liquid and glass tube.
= Density Of liquid.
h= 4cos / e*g*d.
Compressibility and Bulk Modulus
Compressibility: The property by virtue of which fluids undergo a change in volume under the action external pressure is known as compressibility. It decreases with the increase in pressure of fluid.
Bulk modulus of elasticity (K) is defined as the ratio of compressive stress to volumetric strain. Compressibility is the reciprocal of bulk modules of elasticity.
Vapour Pressure:
All liquid has a tendency to evaporate a vaporize.
Molecules are continuously projected from the free surface to the atmosphere.
These ejected molecules are in a gaseous state and exact their own partial vapor pressure on the liquid surface. This pressure is known as the vapours pressure of the liquid (Pv).
Vapour pressure increase with the rise in temperature.
Q8) Explain the Pascal’s law.
A8) The Pascal’s law states as follows:
The Intensity of pressure at any point in a liquid at rest, is the same in all directions.
Proof Let us consider a very small wedge shaped element LMN of a liquid as shown in Fig.
Let px = Intensity of horizontal pressure on the element of liquid.
py= Intensity of vertical pressure on the element of liquid.
pz= Intensity of pressure on the diagonal of the right angle triangles element.
= Angle of the element of the liquid.
Px = Total pressure in the vertical side LN of the liquid.
Py = Total Pressure on the horizontal side MN of the liquid, and
Pz = Total pressure on the diagonal LM of the liquid.
Now Px =pxxLN
and, Py =pyxMN
and, Pz =pzx LM
As the element of the liquid is at rest, therefore the sum of horizontal and vertical components of the liquid pressures must be equal to zero.
Resolving the forces horizontally:
Pz Sin = Px
pz LM Sin = px LN
LM Sin = LN
pz = px
Resolving the forces vertically,
Pz Cos = Py – W
Pz Cos = Py
pz LM cos = py LN
(where W = weight of the liquid element)
Since the element is very small, neglecting its weight, we have
But, LM Cos = MN
pz = py
We get px = py = pz
Hence at any point in a fluid at rest the intensity of pressure is exerted equally in all directions, which is called Pascal’s Law.
Q9) What is buoyancy?
A9) Buoyancy
Whenever a body is immersed wholly or partially in a fluid it is subjected to an upward force which tends to lift it up.
The tendency for an immersed body to be lifted up in the fluid due to an upward force opposite to action of gravity is known as buoyancy.
The force tending to lift up the body under such conditions is known as buoyant force or force of buoyancy.
The magnitude of the buoyant force can be determined by Archimedes principle which states as follows:
“When a body is immersed in a fluid either wholly or partially it is buoyed or lifted up by a force which is equal to the weight of fluid displaced by the body.”
Centre of Buoyancy
The point of application of the force of buoyancy on the body is known as the center of buoyancy. It is always the center of gravity of the volume of fluid displaced.
Meta - Center
It is defined as the point about which a floating body starts oscillating when the body is tilted by a small angle.
The meta-centre may also be defined as the point at which the line of action of the force of buoyancy will meet the normal axis of the body when the body is given a small angular displacement.
Consider a body floating in a liquid as. Let the body is in equilibrium and G is the center gravity and B is the center of Buoyancy.
Let the body is given a small angular displacement in the clockwise direction as shown in fig. the centre of buoyancy, which is the centre of the displaced liquid or centre of gravity of portion of the body submerged in liquid, will be shifted towards right from the normal axis.
Let it is at B1 as shown in Fig.
The line of action of the force of buoyancy in this position, will intersect the normal axis of the body some point say M.
This point M is called Meta-centre.
Meta-Centric Height
The distance MG, the distance between the meta-center of a floating body and the center of gravity of the body is called Meta-centric height.
Q10) Determine the concept of stability of floating body.
A10) The stability of a floating body is determined from the position of Meta-centre (M).
a). Stable Equilibrium
If the point M is above G, the floating body will be in stable equilibrium as shown in Fig.
M is above G(stable equilibrium) M is below G (unstable equilibrium)
If a slight angular displacement is given to the floating body in the clockwise direction, the center of buoyancy shifts from B to B1, such that the vertical line through B1 cuts at M.
Then the buoyant force FB through B1 and weight W through G constitute a couple acting in the anti-clockwise direction and thus bringing the floating body in the original position.
Unstable Equilibrium:
If the point M is below G, the floating body will be in unstable equilibrium as shown in fig.
The disturbing couple is acting in the clockwise direction.
The couple due to buoyant force FB and W is also acting in clock direction and thus overturning the floating body.
Neutral Equilibrium:
If the point M is at the centre of gravity of the body. The floating body will be in neutral equilibrium.
Q11) Define Dimensions of physical quantities and dimensional homogeneity.
A11) Dimensions of physical quantities
Dimensional Homogeneity
P = w h
Dimensions of LHS = ML-1T-2
Dimensions of RHS = ML-2T-2×L = ML-1T-2
Dimensions of LHS = Dimensions of RHS
Hence, equation P= w h is dimensionally homogeneous; so it can be used in any system of units.
Applications of Dimensional Homogeneity:
Q12) Explain Buckingham’s Theorem.
A12)
“If there are n variables in a dimensionally homogeneous equation and if these variables contain fundamental dimensions (such as M, L, T, etc.) then the variables are arranged into (n-m) dimensionless terms.
Mathematically, if any variable X1, depends on independent variables, X2, X3, X4,…… Xn; the functional equation may be written as
-------- 1
Equation 1 can also be written as
-------- 2
It is dimensionally homogeneous equation and contain n variables.
If there are m fundamental dimensions, then according to Buckingham’s Π – theorem, it [eqn] can be written in terms of number of Π – terms in which number of Π terms is equal to (n-m).
Hence equation 2 becomes as
--------3
Each dimensionless 
– term is formed by combining m variables out of the total n variables with one of the remaining (n-m) variable.
These m variables which appear repeated in each of 
– terms are consequently called repeating variables and are chosen from among the variables such that they together involve all the fundamental dimentions and they themselves do not form a dimensionless parameter.
Let in the above case X2, X3 and X4 are the repeating variables if the fundamental dimensions m(M, L, T) = 3, then each term is written as:
.
.
.
-------- 4
Where a1, b1, c1, a2, b2, c2 etc. are the constants, which are determined by considering dimensional homogeneity. These values are substituted in equation 4 and values of 
are obtained. These values of Π’s are substituted in equation 3.
The final general equation for the phenomenon may then be obtained by expressing anyone of the Π – terms as a function of the other as:
Q13) Define Geometric similarity, Kinematic similarity and Dynamic similarity.
A13) Geometric Similarity
For geometric similarity to exist between the model and the prototype, the ratios of corresponding lengths in the model and the prototype must be same and the included angles between the corresponding sides must be the same.
Models which are not geometrically similar are known as geometrically distorted model.
Lm = length of model
Hm = height of model
Dm = diameter of model
Am = area of model
Vm = volume of model
And LP, BP, HP, DP, AP and VP = corresponding values of prototype.
Then, for geometric similarity, we must have the relation:
Where Lr is called the scale ratio or the scale factor:
Kinematic similarity:
Kinematic similarity is the similarity of motion.
If at the corresponding points in the model and in the prototype, the velocity or acceleration ratio are same & velocity or acceleration vectors point in the same direction, the two flows are said to be kinematically similar.
(V1)m = velocity of fluid at point 1 in the model.
(V2)m = velocity of fluid at point 2 in the model,
(a1)m = acceleration of fluid at point 1 in the model
(a2)m = acceleration of fluid at point 2 in the model
And (V1)P, (V2)P, (a1)P (a2)P = corresponding values at the corresponding points of fluid, velocity and acceleration in the prototype.
Then, for kinematic similarity, we must have
Similarly, 
The direction of the velocities in the model and prototype should be same.
The geometric similarity is a pre-requisite for kinematic similarity.
Dynamic Similarity
Dynamic similarity is the similarity of forces. The flows in the model and in prototype are dynamically similar if at all the corresponding points, identical types of forces are parallel and bear the same ratio.
In dynamic similarity, the force polygons of the two flows can be superimposed by change in force scale.
(Fi)m = inertia force at apoint in the model,
(Fv)m = viscous force at the point in the model,
(Fg)m = gravity force at the point in the model
And (Fi)P, (Fv)P, (Fg)P = corresponding values of forces at the corresponding points in prototype.
Then, for dynamic similarity we have
The direction of the corresponding forces at the corresponding points in the model and prototype should also be same.
Q14) Explain the dimensionless parameters.
A14) Reynold’s Number (Re)
It is defined as the ratio of the inertia force to the viscous force.
Reynold’s Number, Re = 
For pipe flow (where the linear dimension is taken as diameter d),
Reynold’s number signifies the relative predominance of the inertia to the viscous forces occurring in the flow systems.
Fraud’s Number (Fr)
It is defined as the square root of the ratio of the inertia force and the gravity force.
Mathematically, 
Froude number governs the dynamics similarity of the flow situations; where gravitational force is more significant and all other forces are comparatively negligible.
Euler’s Number (Eu)
It is defined as the square root of the ratio of the inertia force to the pressure force.
The Euler number is important in the flow problems / situations in which a pressure gradient exists.
Weber’s Number (We)
It is defined as the square root of the ratio of the inertia force to the surface tension force.
Mathematically, 
This number assumes importance in the following flow situation:
i) Flow of blood in veins and arteries
ii) Liquid atomisation, and
iii) Capillary movement of water in soils
Mach Formula (M)
It is defined as the square root of the ratio of the inertia force to the elastic force.
Mathematically, 
The Mach number is important in compressible flow problems at high velocities, such as high velocity flow in pipes or motion of high speed projectiles and missiles.
Q15) Calculate the dynamic viscosity of an oil, which is used for lubrication between a square plate of size 0.8 m x 0.8 m &an inclined plane with angle of inclination 300 as shown in fig. The weight of the square plate is 300 N & it slides down the inclined plane with a uniform velocity of 0.3 m/s. The thickness of oil film is 7.5 mm.
A15) Given
A = 0.8 x 0.8 = 0.64 m2
= 300
W = 300 N
u = 0.3 m/s
dy = 1.5 mm = 1.5 x 10-3 m
1). F = 300Sin = 300 sin30 = 150 N
2). = F/A = 150/0.64 = 234.375 N/m2
3). = du/dy
xdy/du =
234.375 x 1.5 x 10-3/0.3 =
= 1.17 Ns/m2
Q16) The dynamic viscosity of an oil, used for lubrication between a shaft & sleeve is 6 poise. The shaft is of diameter 0.4m & rotates at 190 rpm. Calculate the power lost in the bearings for a sleeve length of 90 mm. The thickness of the oil film is 1.5 mm.
A16) Given
= 6 poise = 6/10 = 0.6 Ns/m2
D = 0.4 m
N = 190 rpm
L = 90 mm = 0.09 m
dy = 1.5 mm = 1.5 x 10-3 m
1). u = πDN/60 = π x 0.4 x 190/60 = 3.98 m/s
2). du = u-0 = 3.98 m/s
3). = du/dy = 0.6 x 3.98/1.5 x 10-3 = 1592 N/m2
4). F = x Area
= 1592 x π x D x L
= 1592 x π x 0.4 x 0.09
= 180.05 N
5). Torque = F x D/2 = 180.05 x 0.4/2 = 36.01 Nm
6). Power lost = 2πNT/60 = 2π x 190 x 36.01/60 = 716.48 W
Q17) Two large plane surfaces are 2.4 cm apart. The space between the surfaces is filled with glycerin. What force is required to drag a very thin plate of surface area as square meter between the two large plane surfaces at a speed of 0.6 m/s if,
i). The thin plate is in the middle of the two plane surfaces.
ii). The thin plate is at a distance of 0.8 cm from one of the plane surfaces?
A17) Take = 8.10 x 10-1 Ns/m2
Given:
A = 0.5 m2
u = 0.6 m/s
= 8.10 x 10-1 Ns/m2
Case 1 When the thin plate is in the middle of the two planes.
dy = 0.024/2 = 0.012 m
= du/dy
= 8.10 x 10-1 x 0.6/0.012 = 40.5 N/m2
F = x A x 2
= 40.5 x 0.5 x 2
= 40.5 N
ii). When the thin plate is at a distance of 0.8 cm from one of the plane.
dy1 = 0.008 m
dy2 = 0.024 – 0.008 = 0.016 m
1 = du/dy1 = 8.10 x 10-1 x 0.6/0.008
= 60.75 N/m2
F1 = 1 x A = 60.75 x 0.5 = 30.375 N
2 = du/dy2 = 8.10 x 10-1 x 0.6/0.016
= 30.375 N/m2
F2 = 2 x A = 30.375 x 0.5 = 15.187 N
F = F1 + F2
= 30.375 + 15.187
= 45.562 N
Q18) A vertical gap 2.2 cm wide of infinite extent contains a fluid of viscosity 2 Ns/m2& sp.gr 0.9. A metallic plate 1.2m x 1.2m x 0.2cm is to be lifted up with a constant velocity of 0.15 m/sec, through the gap. If the plate is in the middle of the gap. Find force required.
A18) The weight of the plate is 40N.
Width of gap = 2.2 cm = 0.022 m
= 2 Ns/m2
S = 0.9
= S x 9810 = 0.9 x 9810
= 8829 N/m3
Size of plate = (1.2 x 1.2 x (2 x 10-3))m
du = 0.15 m/sec
Weight of plate = 40 N
1). Distance of plate from vertical surface of the gap, when plate is in the middle.
dy = 0.022 – 2 x 10-3/2 = 0.01 m
2). = x du/dy = 2 x 0.15/0.01 = 30 N/m2
3). Total shear force = 2 x x A
F = 2 x 30 x (1.2 x 1.2)
= 86.4 N
4). The upward thrust = weight of fluid displaced
= x v
= 8829 x 1.2 x 1.2 x 2 x 10-3
= 25.43 N
5). Effective weight of= weight ofplate – Upward thrust the plate
= 40 – 25.43
= 14.57 N
6). Total force required to = F + 14.57
lift the plate up
= 86.4 + 14.57
= 100.97 N
Q19) A rectangle plane surface is 2 m wide & 3 m deep. It lies in vertical plane in water. Determine the total pressure & position of centre of pressure when its upper edge is horizontal & a) Coincides with water surface
b) 2.5 m below the free water surface.
A19)
Given
b = 2m
d = 3m
a). Upper edge coincides with water surface
A = b x d = 2 x 3 = 6 m2
X = d/2 -3/2 = 1.5 m
Total Pressure
F = AX
= 9810 x 6 x 1.5
= 88290 N
IG = bd3/12 = 2 x 33/12 = 4.5 m4
Centre of Pressure
h = IG/AX + X = 4.5/(6 x 1.5) + 1.5 = 2m
b). Upper edge 2.5 m below water surface
X = 2.5 + 3/2 = 4m
Total pressure F = Ax
= 9810 x 6 x 4
= 235440 N
h = IG/AX + X = 4.5/(6 x 4) + 4 = 4.1875 m
Q20) A heavy car plunges into a lake during an accident & land at the bottom of the lake on its wheel. The door is 1.2 m high & 1m wide, the top edge of the door is 8m below the free surface of the water. Determine total pressure acting on door approximating it as a vertical rectangular plate & centre of pressure.
A20) Given
b = 1 m
d = 1.2 m
A = b x d = 1 x 1.2 = 1.2 m2
X = 8 + d/2 = 8 + 1/2/2 = 8.6 m
F = AX
= 9810 x 1.2 x 8.6
= 101.24 x 103 N
IG = bd3/12 = 1 x 1.23/12 = 0.144 m4
h = IG/AX + X = 0.144/(1.2 x 8.6) + 8.6 = 8.614 m
Q21) A circular plate 1.5 m diameter is submerged in water with its greatest & least depth below the surface being 2 m & 0.75 m respectively. Determine total pressure & centre of pressure.
A21) Given
d = 1.5 m
1). A = π/4 x d2 = π/4 x 1.52 = 1.767 m2
2). X = Greatest depth + Least depth / 2
= (2 + 0.75)/2 = 1.375 m
3). F = AX
= 9810 x 1.767 x 1.375
= 28.83 x 103 N
4). Sin = 1.25/1.5 = 0.833
5). IG = π/64 x d4 = π/64 x 1.54 = 0.248 m4
6). h = IG Sin2/AX + X
0.248 x 0.8332/1.767 x 1.375 + 1.375
= 1.446 m
Q22) A triangular plate of 1m base & 1.5 m altitude is immersed in water. The plate is inclined at 300 with free water surface & base is parallel to & at a depth of 2 m from water surface. Find total pressure & centre of pressure.
A22) Given
b = 1 m
h = 1.5 m
= 300
A = ½ x b x h = ½ x 1 x 1.5 = 0.75 m2
X = 2 + 1.5/3 x Sin 30 = 2.25 m
F = AX
= 9810 x 0.75 x 2.25
= 16.55 x 103 N
IG = bh3/36 = 1 x 1.53/36 = 0.09375 m4
h = IGSin2/AX + X
= 0.09375 x (Sin 30)2/0.75 x 2.25 + 2.25
= 2.264 m
